NO

Problem:
 f(x,x) -> f(i(x),g(g(x)))
 f(x,y) -> x
 g(x) -> i(x)
 f(x,i(x)) -> f(x,x)
 f(i(x),i(g(x))) -> a()

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [i](x0) = x0,
   
   [a] = 4,
   
   [f](x0, x1) = x0 + 2x1 + 4,
   
   [g](x0) = x0
  orientation:
   f(x,x) = 3x + 4 >= 3x + 4 = f(i(x),g(g(x)))
   
   f(x,y) = x + 2y + 4 >= x = x
   
   g(x) = x >= x = i(x)
   
   f(x,i(x)) = 3x + 4 >= 3x + 4 = f(x,x)
   
   f(i(x),i(g(x))) = 3x + 4 >= 4 = a()
  problem:
   f(x,x) -> f(i(x),g(g(x)))
   g(x) -> i(x)
   f(x,i(x)) -> f(x,x)
   f(i(x),i(g(x))) -> a()
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [i](x0) = x0,
    
    [a] = 0,
    
    [f](x0, x1) = 4x0 + x1 + 1,
    
    [g](x0) = x0
   orientation:
    f(x,x) = 5x + 1 >= 5x + 1 = f(i(x),g(g(x)))
    
    g(x) = x >= x = i(x)
    
    f(x,i(x)) = 5x + 1 >= 5x + 1 = f(x,x)
    
    f(i(x),i(g(x))) = 5x + 1 >= 0 = a()
   problem:
    f(x,x) -> f(i(x),g(g(x)))
    g(x) -> i(x)
    f(x,i(x)) -> f(x,x)
   Unfolding Processor:
    loop length: 4
    terms:
     f(x141,x141)
     f(i(x141),g(g(x141)))
     f(i(x141),i(g(x141)))
     f(i(x141),i(i(x141)))
    context: []
    substitution:
     x141 -> i(x141)
    Qed